We are given that an initial investment of $12,000 grew to $56,946.68 over 10 years, and we need to find the annual interest rate, assuming the investment is compounded annually.

The formula for compound interest is:

$A = P(1 + r)^t$

Where:

• $A$ is the final amount (56,946.68),
• $P$ is the principal or initial amount (12,000),
• $r$ is the annual interest rate (the unknown),
• $t$ is the time in years (10 years).

Substitute the values into the formula:

$56,946.68 = 12,000(1 + r)^{10}$

Now, solve for $r$:

$\frac{56,946.68}{12,000} = (1 + r)^{10}$

$4.74555667 = (1 + r)^{10}$

Now take the 10th root of both sides:

$(1 + r) = \sqrt[10]{4.74555667}$

Finally, subtract 1 to isolate $r$:

$r = \sqrt[10]{4.74555667} - 1$

Let me calculate that.The annual interest rate, compounded annually, is approximately 16.85%.

Would you like more details or further explanation?

Here are 5 related questions for further exploration:

1. What would the investment have been worth after 15 years at the same interest rate?
2. How does the formula for compound interest change if interest is compounded monthly instead of annually?
3. What is the future value of an investment of $20,000 after 10 years at a 16.85% annual interest rate?
4. How long would it take for an investment of $12,000 to double at a 16.85% annual interest rate?
5. What would the interest rate be if the same investment grew to $100,000 in 10 years?

Tip: For a quick estimate of doubling time at a constant interest rate, you can use the "Rule of 72", which approximates the doubling time by dividing 72 by the annual interest rate.